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\begin{document}% \tiny \scriptsize \footnotesize \small \normalsize \large \Large \LARGE \huge \Huge 
\begin{center}
{\Huge Zorn's Lemma}\bigskip\\ 
Pierre-Yves Gaillard\footnote{ORCID https://orcid.org/0000-0002-7960-1698} 
\end{center} 

\noi In the text \emph{Zorn's Lemma} 

\noi \href{https://scholar.google.com/scholar?cluster=13609030796755773403&hl=en&as_sdt=0,5}{https://scholar.google.com/scholar?cluster=13609030796755773403\& hl=en\& as\_sdt=0,5} 

\noi Daniel R. Grayson summarizes Kneser's proof of Zorn's Lemma given in

\noi Hellmuth Kneser. Eine direkte Ableitung des Zornschen Lemmas aus dem Auswahlaxiom. \emph{Math. Z.}, 53:110–113, 1950. \href{http://gdz.sub.uni-goettingen.de/dms/load/img/?PID=GDZPPN00238180X}{http://gdz.sub.uni-goettingen.de/dms/load/img/?PID=GDZPPN00238180X}

\noi The purpose of this text is to add more details to Grayson's text. Thank you to Murray Eisenberg for having pointed out Grayson's text to me! 

\begin{thm}[Zorn's Lemma]\label{ZL}
Let $P$ be a poset all of whose well-ordered subsets have an upper bound. Then $P$ has a maximal element. 
\end{thm} 

If $p$ is a member of a poset $P$, we denote by $P_{<p}$ the set of those elements of $P$ which are less than $p$. We say that a subset $D$ of a poset $P$ is \textbf{downward closed} if $P\ni p<d\in D$ implies $p\in D$, that is, if 
\begin{equation}\label{Edd}
P_{<d}=D_{<d}
\end{equation} 
for all $d\in D$. We define that partial order $\preceq$ on the set of all subsets of $P$ by decreeing that $S\preceq T$ if $S$ is downward closed in $T$. 

\begin{prop}\label{P2} 
If $D$ is a proper downward closed subset of a well-ordered set $W$, and if $m=\min(W\setminus D)$, then $D=W_{<m}$. $\square$\footnote{The symbol $\square$ at the end of a statement means that the proof is left to the reader as an easy exercise.} 
\end{prop} 

\begin{prop}\label{P3} 
A union of downward closed subsets is downward closed. $\square$ 
\end{prop} 

\begin{prop}\label{P4} 
If $\mc W$ is a set of well-ordered subsets of $P$, and if $\mc W$ is totally ordered by $\preceq$, then the union $U=\bc_{W\in\mc W}W$ is well-ordered. 
\end{prop} 

\begin{lem}\label{L5} 
In the setting of Proposition~\ref{P4} each $W\in\mc W$ is downward closed in $U$. 
\end{lem} 

\begin{lem}\label{L6} 
If $A$ and $D$ are subsets of a poset $P$, and if $D$ is downward closed in $P$, then $A\cap D$ is downward closed in $A$. $\square$ 
\end{lem} 

To prove Theorem~\ref{ZL} we assume that $P$ has no maximal element and we try to get a contradiction.  

For any well-ordered subset $W$ of $P$ there is an element $p(W)$ in $P_{>W}$, that is, an element $p(W)$ in $P$ such that $p(W)>w$ for all $w$ in $W$. Indeed, $W$ has an upper bound $b$, and, $b$ being not maximal, there is a $p(W)$ such that $p(W)>b$. 

Let $\mathcal W$ be the set of those well-ordered subsets $W$ of $P$ such that 
$$
p(W_{<w})=w
$$ 
for all $w$ in $W$. If $W$ is in $\mathcal W$, then, clearly, so is $W\cup\{p(W)\}$. Thus $\mathcal W$ (ordered by inclusion) has no maximal element. We will reach the contradiction we are after by showing that $\mathcal W$ does have a maximal element, namely the set $U:=\bc_{W\in\mathcal W}W$. Hence, Theorem~\ref{ZL} will follow from 

\begin{prop}\label{P7} 
The set $U$ is in $\mathcal W$.  
\end{prop} 

\begin{lem}\label{L8} 
A downward closed subset of a member of $\mc W$ is in $\mc W$. $\square$ 
\end{lem} 

\begin{lem}\label{L9} 
If $W,X\in\mc W$ and $W\prec X$, then we have $p(W)=\min(X\sm W)$ and $W=X_{<p(W)}$. 
\end{lem} 

\begin{lem}\label{L10} 
The set $\mc W$ satisfies the assumptions of Proposition~\ref{P4}. 
\end{lem} 

We now prove the seven remaining statements. 

\begin{proof}[Proof of Lemma~\ref{L5}] 
Let $U\ni u<w\in W\in\mc W$, and let us show $u\in W$. We have $u\in X\in\mc W$ for some $X$, and we get $u\in W$ because $W\preceq W\cup X$. 
%If $W$ is in $\mc W$, then $U$ is also the union of the sets $X\in\mc W$ such that $X\succeq W$, and $U\ni u<w\in W$ implies $u\in X$ for some $X\succeq W$, hence $u\in W$. 
%Let $U\ni u<w\in W\in\mc W$, and let us show $u\in W$. We have $u\in X\in\mc W$ for some $X$, and we get $W\cup X\ni u<w\in W\preceq W\cup X$, hence $u\in W$. 
\end{proof} 

\begin{proof}[Proof of Proposition~\ref{P4}]
Let $A$ be a nonempty subset of $U$. There is a $W$ in $\mc W$ such that $W\cap A\ne\varnothing$, and Lemmas \ref{L5} and \ref{L6} imply $\min(W\cap A)=\min(A)$. 
\end{proof} 

\begin{proof}[Proof of Lemma~\ref{L8}]
Follows from \eqref{Edd}. 
\end{proof} 

\begin{proof}[Proof of Lemma~\ref{L9}]
Follows from Proposition~\ref{P2}. 
\end{proof} 

\begin{proof}[Proof of Lemma~\ref{L10}]
Let $W,X\in\mc W$. It suffices to show that we have $W\preceq X$ or $X\preceq W$. Assume by contradiction that $W$ and $X$ are incomparable. Let $\mc D_W$ be the set of all downward closed subsets of $W$, and define $\mc D_X$ similarly. By Lemma~\ref{L8} the sets $\mc D_W$ and $\mc D_X$ are contained in $\mc W$. Put $D:=\bc_{E\in\mc D_W\cap\mc D_X}E$. Then Proposition~\ref{P3} implies $D\in\mc D_W\cap\mc D_X$. We also have $D\prec W$ and $D\prec X$. Set $w:=\min(W\sm D)$ and $x:=\min(X\sm D)$. Lemma~\ref{L9} yields $w=p(D)=x$, and $D\cup\{p(D)\}\in D_W\cap\mc D_X$, contradiction. 
\end{proof} 

\begin{proof}[Proof of Proposition~\ref{P7}]
Lemma~\ref{L10} and Proposition~\ref{P4} imply that $U$ is well-ordered. It only remains to show $p(U_{<u})=u$ for $u\in U$. Since $p(W_{<u})=u$, it suffices to prove $W_{<u}=U_{<u}$. In view of \eqref{Edd} this follows from Lemmas \ref{L10} and \ref{L5}. 
\end{proof} 

\begin{proof}[Proof of Theorem~\ref{ZL}]
As already mentioned, the theorem follows from Proposition~\ref{P7}. 
\end{proof} 

\end{document}